Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool cmp(pair<long long, int> p1, pair<long long, int> p2) {
  return p1.first < p2.first;
}
long double n, h;
long double areaa(long double mid) {
  long double x = mid / h;
  x /= 2;
  long double h1 = sqrt(mid * mid - x * x);
  return h1 * mid * 0.5;
}
int main() {
  cin >> n >> h;
  long double h1 = sqrt(h * h - 0.25);
  long double area = h1 * h * 0.5;
  long double pice = area / n;
  long double first = 0, tot = 0;
  for (int i = 1; i < n; i++) {
    long double st = 0.0, end = h, mid;
    for (int j = 0; j < 1000; j++) {
      mid = (st + end) * 0.5;
      long double ss = areaa(mid);
      if ((ss - tot) < pice)
        st = mid;
      else
        end = mid;
    }
    cout << fixed << setprecision(12);
    cout << st << " ";
    tot = areaa(mid);
  }
  return 0;
}